 Solution

An algebraic element $\alpha$ over $K$ is a root of a nonzero polynomial in $K[T]$. Its degree is the degree of its minimal polynomial, equivalently $[K(\alpha ):K]$. Since $\alpha$ satisfies $T^2-{\alpha }^2$ over $K({\alpha }^2)$, the tower law gives $[K(\alpha ):K({\alpha }^2)]\le 2$ and makes that degree divide the odd integer $[K(\alpha ):K]$. It must therefore be one, proving $\boxed{K(\alpha )=K({\alpha }^2)}$.

For the addition assertion, let polynomial relations of degrees $r,s$ express ${\alpha }^r$ and ${\beta }^s$ in terms of lower powers. The finite-dimensional $K$-space spanned by ${\alpha }^i{\beta }^j$ for $0\le i<r$, $0\le j<s$ contains one and is closed under multiplication by $\alpha +\beta$. Thus $1,\alpha +\beta ,\dots ,(\alpha +\beta {)}^{rs}$ are linearly dependent. Their dependence is a nonzero polynomial relation over $K$, so $\alpha +\beta$ is algebraic.